2020 AMC 10B Problem 24

1. Name the integer. $\lfloor \sqrt n \rfloor$ is a whole number, so call it $k$. Then $\frac{n + 1000}{70} = k$, so
$$n = 70k - 1000.$$
2. What "$\lfloor \sqrt n \rfloor = k$" means. It means $\sqrt n$ is at least $k$ but less than $k + 1$. Squaring, $n$ is at least $k^2$ but less than $(k+1)^2$:
$$k^2 \le 70k - 1000 < (k + 1)^2.$$
These are two conditions, and we need both.
3. First condition: $k^2 \le 70k - 1000$, that is, $k^2 - 70k + 1000 \le 0$. This factors as $(k - 20)(k - 50) \le 0$, which holds exactly for $20 \le k \le 50$.
4. Second condition: $70k - 1000 < k^2 + 2k + 1$, that is, $k^2 - 68k + 1001 > 0$. Rather than solve it exactly, test values near the ends of $20$ to $50$:

1. Why the middle fails. $n$ goes up by $70$ each time $k$ goes up by $1$, while $(k+1)^2$ goes up by $2k + 3$. So the gap $(k+1)^2 - n$ keeps shrinking while $2k + 3 < 70$ and keeps growing after that. It is negative from $k = 22$ to $k = 46$ (both ends checked in the table) and positive outside. (Solving $k^2 - 68k + 1001 = 0$ gives $k \approx 21.6$ and $k \approx 46.4$, which agrees.)
2. Collect. $k = 20, 21, 47, 48, 49, 50$: six values, each giving a different positive $n$: $400, 470, 2290, 2360, 2430, 2500$.

The answer is $\boxed{\textbf{(C)}\ 6}$.

An equation with a floor is really two inequalities: at least $k^2$, and less than $(k+1)^2$. Naming the whole-number value $k$ makes $n$ a simple expression in $k$. The straight line $70k - 1000$ climbs at a steady rate while the squares speed up, so the solutions sit in two short groups at the ends, not one long run. The choices $30$ and $32$ are there to catch anyone who checks only the first inequality.

The right side is an integer, so let $k = \lfloor \sqrt n \rfloor = \dfrac{n + 1000}{70}$. Then $n = 70k - 1000$, and the floor condition $k \le \sqrt n < k + 1$ becomes
$$k^2 \le 70k - 1000 < (k+1)^2.$$

Left inequality: $k^2 - 70k + 1000 \le 0$. The roots of $k^2 - 70k + 1000$ are $k = \frac{70 \pm \sqrt{4900 - 4000}}{2} = \frac{70 \pm 30}{2} = 20, 50$, so $20 \le k \le 50$.

Right inequality: $70k - 1000 < k^2 + 2k + 1$, i.e. $k^2 - 68k + 1001 > 0$. The roots are $k = \frac{68 \pm \sqrt{4624 - 4004}}{2} = 34 \pm \frac{\sqrt{620}}{2}$. Since $\sqrt{620} \approx 24.9$, the roots are about $21.55$ and $46.45$, so we need $k \le 21$ or $k \ge 47$.

Combining: $k \in \{20, 21, 47, 48, 49, 50\}$, six values. Each gives a positive $n = 70k - 1000$ (namely $400, 470, 2290, 2360, 2430, 2500$), and distinct $k$ give distinct $n$.

Spot check: $k = 21$ gives $n = 470$, $\sqrt{470} \approx 21.7$, floor $21$. Good. $k = 22$ gives $n = 540$, $\sqrt{540} \approx 23.2$, floor $23 \ne 22$. Fails, as predicted.

The answer is $\boxed{\textbf{(C)}\ 6}$.